## The Lie Algebra of a Lie Group

Since a Lie group is a smooth manifold we know that the collection of vector fields form a Lie algebra. But this is a big, messy object because smoothness isn’t a very stringent requirement on a vector field. The value can’t vary too wildly from point to point, but it can still flop around a huge amount. What we really want is something more tightly controlled, and hopefully something related to the algebraic structure on to boot.

To this end, we consider the “left-invariant” vector fields on . A vector field is left-invariant if the diffeomorphism of left-translation intertwines with itself for all . That is, must satisfy ; or to put it another way: . This is a very strong condition indeed, for any left-invariant vector field is determined by its value at the identity . Just set and find that

The really essential thing for our purposes is that left-invariant vector fields form a Lie subalgebra. That is, if and are left-invariant vector fields, then so is their sum , scalar multiples — where is a constant and not a function varying as we move around — and their bracket . And indeed left-invariance of sums and scalar multiples are obvious, using the formula and the fact that is linear on individual tangent spaces. As for brackets, this follows from the lemma we proved when we first discussed maps intertwining vector fields.

So given a Lie group we get a Lie algebra we’ll write as . In general Lie groups are written with capital Roman letters, while their corresponding Lie algebras are written with corresponding lowercase fraktur letters. When has dimension , also has dimension — this time as a vector space — since each vector field in is uniquely determined by a single vector in .

We should keep in mind that while is canonically isomorphic to as a vector space, the Lie algebra structure comes not from that tangent space itself, but from the way left-invariant vector fields interact with each other.

And of course there’s the glaring asymmetry that we’ve chosen left-invariant vector fields instead of right-invariant vector fields. Indeed, we could have set everything up in terms of right-invariant vector fields and the right-translation diffeomorphism . But it turns out that the inversion diffeomorphism interchanges left- and right-invariant vector fields, and so we end up in the same place anyway.

How does the inversion act on vector fields? We recognize that , and find that it sends the vector field to . Now if is left-invariant then for all . We can then calculate

where the identities and reflect the simple group equations and , respectively. Thus we conclude that if is left-invariant then is right-invariant. The proof of the converse is similar.

The one thing that’s left is proving that if and are left-invariant then their right-invariant images have the same bracket. This will follow from the fact that , but rather than prove this now we’ll just push ahead and use left-invariant vector fields.

“we get a Lie algebra we’ll write as \mathfrak{G}” – this should be lowercase, given the context.

Comment by Andrei | June 8, 2011 |

Sorry, typo. Fixed.

Comment by John Armstrong | June 8, 2011 |

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